Incredible vanilla geometry again

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anantmudgal09

1979 posts

anantmudgal09

#1 h Mar 7, 2021, 10:32 AM • 5 Y

Y by Aritra12, p_square, MrOreoJuice, parmenides51, Rounak_iitr

In a convex quadrilateral $ABCD$ , $\angle ABD=30^\circ$ , $\angle BCA=75^\circ$ , $\angle ACD=25^\circ$ and $CD=CB$ . Extend $CB$ to meet the circumcircle of triangle $DAC$ at $E$ . Prove that $CE=BD$ .

Proposed by BJ Venkatachala

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anantmudgal09

1979 posts

anantmudgal09

#2 h Mar 7, 2021, 11:08 AM • 7 Y

Y by Aimingformygoal, nikaaryu, Sumitrajput0271, Aritra12, nnathan, primesarespecial, Pratik12

I will post the official solution:

Solution

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Reason: forgot hide tags

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L567

1184 posts

L567

#3 h Mar 7, 2021, 11:11 AM • 2 Y

Y by Pluto1708, nathantareep

Nice one!

Let $F$ be the point on $BC$ such that $CF = CB = CD$ . This actually makes $CFD$ and equilateral triangle and so now a bit of angle chase to show that $\angle FAC = \angle FCA = 35^\circ$ and so $FA = FC$ and now some more angle chase to get that $\angle DAC = 30^\circ$ . Then, just use law of sines to get that $CE = BD$

This post has been edited 3 times. Last edited by L567, Sep 25, 2021, 11:59 AM

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Bahnhofstrasse

25 posts

Bahnhofstrasse

#4 h Mar 7, 2021, 11:27 AM • 1 Y

Y by Pluto1708

Proceeds to Trig Bash

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554183

484 posts

554183

#5 h Mar 7, 2021, 11:32 AM

Y by

Redxated

This post has been edited 2 times. Last edited by 554183, Jun 13, 2021, 8:34 AM

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Bahnhofstrasse

25 posts

Bahnhofstrasse

#6 h Mar 7, 2021, 11:50 AM • 1 Y

Y by Prabh2005

Cheers @above

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N1RAV

160 posts

N1RAV

#7 h Mar 7, 2021, 12:08 PM

Y by

@anantmudgal09,

Is there a way by which we can draw the figure only with a pencil, straightedge, a compass?

This is because I don't know how to draw an angle of $25^{\circ}$

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Math_god__

5 posts

Math_god__

#8 h Mar 7, 2021, 12:15 PM

Y by

yeah @2above I bashed too...

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Nuterrow

254 posts

Nuterrow

#9 h Mar 7, 2021, 12:39 PM • 2 Y

Y by Mango247, Mango247

I was simplifying my trig bash somehow by Ptolemy and random similarities...I saw the examiner taking other's paper...luckily I was in the very last seat and somehow fakesolved xd

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Megamind123

88 posts

Megamind123

#10 h Mar 7, 2021, 1:07 PM • 4 Y

Y by Pluto1708, Mango247, Mango247, Mango247

N1RAV wrote:

@anantmudgal09,

Is there a way by which we can draw the figure only with a pencil, straightedge, a compass?

This is because I don't know how to draw an angle of $25^{\circ}$

There is a method to approx a 20 degree angle (it gives 19.1 degree which you can use)

Search google/YouTube for it

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MeineMeinung

68 posts

MeineMeinung

#11 h Mar 7, 2021, 1:28 PM

Y by

Let the perpendicular bisector of $BD$ intersect $AB$ at $F$ . We get
1. $\angle FCD = \frac{1}{2} \angle BCD = 50^{\circ}$ and since $\angle ACD = 25^{\circ}$ , $CA$ bisects $\angle FCD$ .
2. $\angle BFC = 180^{\circ} - \angle FBC - \angle FCB = 60^{\circ}$ . Due to symmetry, $\angle CFD = \angle BFC = 60^{\circ}$ . Therefore, $AF$ must be the external angle bisector of $\angle DFC$ .

Now, observe $\triangle CFD$ . Notice that $A$ is the intersection of internal angle bisector of $\angle FCD$ and external angle bisector $\angle CFD$ and so $A$ must be $C$ -excenter of $\triangle CFD$ . So, $AD$ must be the external angle bisector of $\angle CDF$ and hence $\angle FDA = 90^{\circ} - \frac{1}{2} \angle FDC = 90^{\circ} - \frac{1}{2} \angle FBC = 55^{\circ}$ . Therefore, $\angle CDA = \angle CDF + \angle FDA = 125^{\circ}$ .

Now, with a little more angle chase, we can find that $\angle CED = \angle CAD = 30^{\circ}$ and $\angle EDC = 50^{\circ}$ . By law of sine on $\triangle ECD$ , we get $\frac{EC}{CD} = \frac{\sin 50^{\circ}}{\sin 30^{\circ}} = 2\sin 50^{\circ}$ . But from isosceles triangle $\triangle BCD$ , it is clear that $\frac{BD}{2 BC} = \sin 50^{\circ}$ . Since $BC = CD$ , we conclude that $BD = EC$ .

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Piyushkhattar

3 posts

Piyushkhattar

#12 h Mar 7, 2021, 1:36 PM

Y by

Pls somebody send figure fir the prblem

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L567

1184 posts

L567

#13 h Mar 7, 2021, 1:46 PM • 1 Y

Y by p_square

Piyushkhattar wrote:

Pls somebody send figure fir the prblem

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i3435

1349 posts

i3435

#14 h Mar 7, 2021, 2:00 PM • 4 Y

Y by Aimingformygoal, Mango247, Mango247, Mango247

My Sol

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554183

484 posts

554183

#15 h Mar 7, 2021, 2:33 PM

Y by

My trig bash

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kapilpavase

595 posts

kapilpavase

#34 h Mar 9, 2021, 7:29 AM • 26 Y

Y by 554183, MrOreoJuice, p_square, Nuterrow, NumberX, 636510, Rg230403, Aimingformygoal, anantmudgal09, Siddharth03, N1RAV, Kayak, gabrupro, Pluto1708, Maths_is_my_life, Math_olympics, TinTin028, spicemax, NJOY, CaptainLevi16, Hamroldt, ayan_mathematics_king, OlympusHero, Jc426, rama1728, ATGY

Let $X$ be such that $BDX$ is equilateral.Then $BA\perp XD$ means $XAD$ is isosceles. Also since $CA$ bisects $\angle{XCD},XACD$ is cyclic. Also easily seen $XE\parallel CD$ . Hence $CE=XD=BD$ .
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Edit: Lol. solution of almost same size of the problem

This post has been edited 4 times. Last edited by kapilpavase, Mar 9, 2021, 8:23 AM

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Abhinavsrijan729

4 posts

Abhinavsrijan729

#35 h Mar 10, 2021, 5:18 AM

Y by

@anantmudgal09
@anyone
I initially chased angles and wrote some of them mentioned above , but then , I dived to trigonometry and simplified it to a good extent and in last it was just left to prove that the small expression consisting of some weird trigo ratios is equal to 1 ,Can I hop partial marks for it(how much ) , bcoz I read somewhere that no partial marks are given for partially solved trigo approach in INMO

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Agsh2005

70 posts

Agsh2005

#36 h Mar 11, 2021, 6:29 PM • 1 Y

Y by Rania06

Call $\angle ADC = \theta$
We are going to find out $\theta$ via trigonometry.
Let $CB=CD=x$
We obtain $BD= 2CBsin(50^{\circ})$ (Cosine Rule)
By Sine Rule in $\Delta ADB$
$AD= \frac{xsin(50^\circ)}{sin(35^{\circ}+ \theta)}$
Sine Rule (again...) in $\Delta ADC$ and combining above we get
$sin(35+\theta)= 2cos(25^\circ)sin(\theta)\dots 1$
$\angle ADB= 115^{\circ} - \theta \implies \theta \leq 115^{\circ}\dots 2$
Also from 1 we have $sin^2(\theta)= \frac{sin^2(35^{\circ})}{(2sin(65^\circ)-cos(35^\circ))^2 +sin^2(35^\circ)}= \frac{1}{4} \implies \theta= 30^{\circ}$

Now define $B'= DB \cap (DAC)$ Now, in $\Delta CB'E$ we find that $\angle CB'E=100^\circ +\theta$

This gives $CE= \frac{x sin(10^{\circ}+\theta)}{sin(\theta)}= 2xsin(50^\circ) = BD$ and we are done!

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sunken rock

4372 posts

sunken rock

#37 h Mar 11, 2021, 8:17 PM • 2 Y

Y by MrOreoJuice, GeoKing

Let $F$ onto $AC$ so that $\angle AFD=30^\circ$ and $O$ reflection of $D$ across $AC$ ; $\triangle CDO$ is equilateral. Clearly $BCDO$ is a kite, so $BO=DO=OF$ , consequently, as $\angle BOC=\angle OBC=65^\circ$ , we get $\angle BOF=70^\circ$ , i.e. $O$ is the circumcenter of $\triangle ABF$ . Additionally, as $\angle AFD=\angle ABD=30^\circ, ABFD$ is cyclic. From $\angle ABF=\angle ABO+\angle OBF=60^\circ$ , we infer $\angle DAF=\angle DBF=30^\circ$ .
Therefore $\angle CED=\angle CAD=30^\circ$ . If $M$ is midpoint of $BD$ and $N$ projection of $C$ onto $DE$ , then $\triangle CDM\cong\triangle DCN$ , hence $DM=CN$ , but $BD=2DM=2CN=CE$ , done.

Best regards,
sunken rock

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W.R.O.N.G

41 posts

W.R.O.N.G

#38 h Mar 20, 2021, 8:28 AM

Y by

Solution

This post has been edited 1 time. Last edited by W.R.O.N.G, Mar 20, 2021, 8:29 AM
Reason: brrrr

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Ali3085

214 posts

Ali3085

#39 h Mar 20, 2021, 11:19 AM • 1 Y

Y by Muaaz.SY

choose $O$ on $AB$ such that $\triangle COD$ is equilateral
so $\angle DOA=50$ so $O$ is the circumcenter of $\triangle CDA$
$\angle CEO =40=\angle CDB$
so $\triangle OCE =\triangle BCD$ so $EC = DB$
and we win

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kapilpavase

595 posts

kapilpavase

#40 h Mar 21, 2021, 4:57 AM • 5 Y

Y by Hamroldt, MatBoy-123, Jc426, N1RAV, p_square

Here is a master method to solve such problems. The idea is to inscribe the configuration in a regular n-gon. We consider $36$ -gon in this case with vertices $X_1,\cdots X_{36}$ . Let $C=X_1$ and $D=X_9$ . Then we observe that $B=X_{29}$ , $X_{15}$ lies on $BA$ and $X_{14}$ lies on $CA$ . Wishfully thinking, we want to prove $\angle{ADB}=85$ , which is equivalent to $X_{12}$ lying on $DA$ . Hence the problem is equivalent to
$\text{In a 36-gon with vertices }X_1,\cdots X_{36}\text{ we have } X_1X_{14},X_9X_{12},X_{29}X_{15} \text{ are concurrent}$ Which is straightforward by complex numbers. You might need the fact that 36th cyclotomic polynomial is $x^{12}-x^6+1$ .

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MatBoy-123

396 posts

MatBoy-123

#42 h Jun 5, 2021, 9:56 AM

Y by

kapilpavase wrote:

Here is a master method to solve such problems. The idea is to inscribe the configuration in a regular n-gon. We consider $36$ -gon in this case with vertices $X_1,\cdots X_{36}$ . Let $C=X_1$ and $D=X_9$ . Then we observe that $B=X_{29}$ , $X_{15}$ lies on $BA$ and $X_{14}$ lies on $CA$ . Wishfully thinking, we want to prove $\angle{ADB}=85$ , which is equivalent to $X_{12}$ lying on $DA$ . Hence the problem is equivalent to
$\text{In a 36-gon with vertices }X_1,\cdots X_{36}\text{ we have } X_1X_{14},X_9X_{12},X_{29}X_{15} \text{ are concurrent}$ Which is straightforward by complex numbers. You might need the fact that 36th cyclotomic polynomial is $x^{12}-x^6+1$ .

Looking Amazing !! From where we can learn about this technique ??

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xeroxia

1133 posts

xeroxia

#43 h Jun 6, 2021, 8:09 AM

Y by

Problem 1: A well-known problem
Let $BCD$ be a triangle with $BC=CD$ and $\angle BCD = 100^\circ$ .
Let $E$ be a point on the ray $[CB$ but outside of segment $[BC]$ such that $\angle CED = 30^\circ$ . Show that $BD = EC$ .

Proof of Problem 1

Back to the original problem.
To solve the problem, we will show $CAD = 30^\circ$ . Since $EADC$ is cyclic, $\angle CED = \angle CAD = 30^\circ$ . So this will take us to the above well-known problem.

In fact it is a more general problem..
Problem 2:
Let $ABCD$ be a quadrilateral such that $\angle ABD = 30^\circ$ , $\angle BCA = 90^\circ - 3\alpha$ , $\angle ACD = 30^\circ - \alpha$ and $CD = CB$ . Show that $\angle CAD = 30^\circ$ .

For $\alpha = 5^\circ$ , it is the same quadrilateral in the original problem.

This type of problem can be easily converted to P inside ABC - Trigonometric Ceva Models. In fact, it belongs to the family - Model 1.

However, I will give a proof to the general quadrilateral version - Problem 2.
Proof of Problem 2

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MV2

126 posts

MV2

#44 h Feb 28, 2022, 5:57 AM

Y by

Trigonometry-

Let $a = BC = CD$ and $\angle DAC = \theta$ .

Claim 1: $\theta = 30^{\circ}$
Proof: Consider $\triangle BDC$ . It is easy to see through cosine rule that $BD = 2a\cos(100^{\circ})$ (1). Consider $\triangle ABD$ . Through angle chasing we see that $\angle ABD = 115^{\circ} - \theta$ . Join $A$ with $E$ to get a cyclic quadrilateral $ADCE$ . Since sum opposite sides of a cyclic quadrilateral sum up to $180^{\circ}$ , we have that, $\angle AEC = 25^{\circ} + \theta$ . Now, consider $\triangle ABE$ . Through angle chasing we see that $\angle BAE = 45^{\circ} = \theta$ . Now, consider $\triangle ACB$ . Using sine rule we see that $AC = a\frac{\sin 70^{\circ}}{\sin 35^{\circ}} \Rightarrow AC = 2a \sin 55^{\circ}$ . Now, consider $\triangle ADC$ . Using sine rule we see that, $\frac{a}{\sin \theta} = \frac{AC}{\sin (155^{\circ} - \theta)} \Rightarrow$ $\frac{a}{\sin \theta} = \frac{2a \sin 55^{\circ}}{\sin (25^{\circ} + \theta)} \Rightarrow \sin (25^{\circ} + \theta) = 2 \sin 55^{\circ} \sin \theta \Rightarrow \cos (65^{\circ} - \theta) = 2 \sin 55^{\circ} \sin \theta \Rightarrow \cos (65^{\circ} - \theta) = \cos (55^{\circ} - \theta) - \cos (55^{\circ} + \theta) \Rightarrow \cos (65^{\circ} - \theta) + \cos (55^{\circ} + \theta) = \cos (55^{\circ} - \theta)$
Using addition to product formula of cosine we have, $\cos (5^{\circ} - \theta) = \cos (55^{\circ} - \theta)$ and $\theta \leq 115^{\circ}$ we see that
$\cos (\theta - 5^{\circ} ) = \cos (55^{\circ} - \theta) \Rightarrow \theta = 30^{\circ}$ .

Claim 2: $DB = CE$
Proof: Finally, consider $\triangle ACE$ . Using sine rule we see that, $CE = AC\frac{\sin 50^{\circ}}{\sin 55^{\circ}}$ (Substituting the value of $\theta = 30^{\circ}$ ). And, $CE = 2a \sin 55^{\circ}\frac{\sin 50^{\circ}}{\sin 55^{\circ}} = 2a \sin 50^{\circ}$ comparing with (1) we have the desired claim. $\blacksquare$

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Mahdi_Mashayekhi

689 posts

Mahdi_Mashayekhi

#45 h Mar 5, 2022, 8:42 AM

Y by

Easy
Let $S$ be on $BA$ such that $\angle ACS = \angle 35$ and $\angle BCS = \angle 40$ . Now we have $BSC$ and $CAS$ are isosceles so $DC = BC = SC = SA$ so $C$ is center of circle $BSD$ so $\angle SDB = \angle SCB/2 = \angle 20$ so $\angle SDC = \angle 60 = \angle SCD$ so $DCS$ is equilateral. $SA = SD = SC$ so $S$ is center of $ADC$ . $SE = SC = CB = CD$ and $\angle ESC = \angle 100 = \angle BCD$ so triangles $ESC$ and $BCD$ are congruent so $BD = EC$ .
we're Done.

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SatisfiedMagma

450 posts

SatisfiedMagma

#46 h Nov 8, 2022, 8:09 AM • 1 Y

Y by Rounak_iitr

Welcome to the dark-side of geometry. Here is a trigonometric solution which is possible to find in exam condition?

Solution: Let $X$ be circumcenter of $\odot(DAC)$ . We will show that $X$ lies on the line $BA$ which is the main part of the problem.
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Observe that since $\triangle BCD$ is isosceles, we have $\angle DBC = 40 ^\circ$ . Finally apply angle sum property in $\triangle ABC$ to get $\angle CAB = 35 ^\circ$ . Call $\angle DAC = \theta$ . We will use trigonometry now to prove $\theta = 30^\circ$ . Apply Law of Sines in $\triangle ACD$ to get
$\begin{align} \frac{AD}{\sin(25^\circ)} & = \frac{CD}{\sin(\theta)} \tag{1} \end{align}$ Apply Law of Sines in $\triangle BAD$ using the fact that $BD = 2\cdot CD\cdot \cos(40^\circ)$ to get
$\begin{align} \frac{AD}{\sin(30^\circ)} & = \frac{2\cdot CD\cdot\cos(40^\circ)}{\sin(35^\circ + \theta)} \tag{2} \end{align}$ Now eliminate $AD$ and $CD$ from $(1)$ and $(2)$ by dividing them. This would give
$\frac{\sin(25^\circ)}{\sin(\theta)} = \frac{\cos(40^\circ)}{\sin(35^\circ + \theta)} = \frac{\sin(50^\circ)}{\sin(35^\circ + \theta)}$ From here on, it is just a brute force calculation. Write $\sin(50^\circ) = 2\sin(25^\circ)\cos(25^\circ)$ and expand $\sin(35^\circ + \theta)$ to get
$2\sin(\theta)\cos(25^\circ) = \sin(35^\circ)\cos(\theta) + \sin(\theta)\cos(35^\circ)$ Divide the equation by $\sin(\theta)$ and solve for $\cot(\theta)$ to finally get
$\begin{align} \cot(\theta) = \frac{2\cos(25^\circ)-\cos(35^\circ)}{\sin(35^\circ)} \tag{3} \end{align}$ Now we prove the following lemma.

Lemma: $\displaystyle\frac{2\cos(25^\circ)-\cos(35^\circ)}{\sin(35^\circ)} = \sqrt{3}$ .

Proof: The idea is to apply sum to product formula once and then write everything in terms of $30^\circ$ and $5^\circ$ angles and force some cancellations.
$\begin{align*} \frac{2\cos(25^\circ)-\cos(35^\circ)}{\sin(35^\circ)} & = \frac{\cos(25^\circ)+(\cos(25^\circ)-\cos(35^\circ))}{\sin(35^\circ)} \\ & = \frac{\cos(25^\circ)+\sin(5^\circ)}{\sin(35^\circ)} \\ & = \frac{\cos(30^\circ - 5^\circ)+\sin(5^\circ)}{\sin(30^\circ-5^\circ)} \\ & = \frac{\sqrt{3}\cos(5^\circ)+3\sin(5^\circ)}{\cos(5^\circ)+\sqrt{3}\sin(5^\circ)} \\ & = \sqrt{3} \end{align*}$ and the lemma is proven. $\square$

Apply the proven Lemma in $(3)$ to get $\cot(\theta) = \sqrt{3} \implies \theta = 30^\circ$ since $0^\circ < \theta < 180^\circ$ . Its about time we get back to geometry.

Re-define $X$ to the intersection of perpendicular bisector of $\overline{AD}$ and $AB$ . Since $\angle DAX = 55^\circ$ we get $\angle AFD = 50^\circ$ . By inscribed angle theorem, $X$ must be the circumcenter of $\odot(EDCA)$ restoring the original definition.

Applying inscribed angle theorem, we can get that $\triangle XCD$ is equilateral. Due to the equilateral triangle, we would have
$\overline{XE} = \overline{XC} = \overline{XD} = \overline{DC} = \overline{BC}$ This would even give $\angle XEC = \angle XCE = 40^\circ$ . Due to all this, one can conclude $\triangle CDB \cong \triangle XEC$ with AAS congruency criteria. It will give $\overline{BD} = \overline{CE}$ and we are done. $\blacksquare$

This post has been edited 2 times. Last edited by SatisfiedMagma, Nov 10, 2022, 8:06 AM
Reason: small asy fix again :yaw:

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Master_of_Aops

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Master_of_Aops

#48 h Jan 4, 2023, 4:39 PM

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Let $\angle DAC = \theta$ and therefore $\angle BDA = 115 - \theta$ . By quadrilateral Ceva, we have that: $\frac{\sin \theta}{\sin(115 - \theta)} = \frac{\sin35}{\sin30} \frac{\sin40}{\sin75} \frac{\sin25}{\sin40}=\frac{\sin25 \sin35 \sin85}{\sin75 \sin30 \sin85} = \frac{\left(\frac{\sin75}{4}\right)}{\sin75 \sin30 \sin85} = \frac{\sin30}{\sin85}$
where the last step is due to the identity: $\sin(60 - x)\sin(x)\sin(60 + x) = \frac{\sin(3x)}{4}$ .
Thus we cleanly get $\theta = 30^\circ$ and now by sine rule, we have that: $BD = CD \cdot \frac{\sin100}{\sin40} = CD \cdot \frac{2\sin50\cos50}{\sin40} = CD \cdot \frac{\sin50}{\sin30} = CE$
And done!

P.S. refer to diagram above

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L13832

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L13832

#49 h Jul 10, 2024, 7:09 PM

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solution from 2 yrs ago with details missing
solution after redoing it

Remark

This post has been edited 4 times. Last edited by L13832, Dec 22, 2024, 6:11 PM

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kes0716

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kes0716

#50 h 5 hours ago

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Choose $F$ (on the opposite side of $C$ with respect to $BD$ ) such that $\Delta FBD$ is equilateral. Because $2\angle ABD=\angle FBD$ , $AB$ is the perpendicular bisector of $FD$ . $AD=AF$ , $AC$ is internal bisector of $\angle FCD$ and $CF>\frac{\sqrt{3}}{2}BD>\frac{1}{2}BD\sec 40^{\circ}=CD$ , so $F, A, C, D$ are cyclic. $\angle EDC=\angle FDC-\angle EDF=(60^{\circ}+40^{\circ})-\angle BCF=50^{\circ}=\angle FCD$ , hence $EC=FD=BD$ .

This post has been edited 2 times. Last edited by kes0716, 5 hours ago

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